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The Helmholtz function $F$ is defined:

$$F = U - TS$$

$$\implies dF = dU - TdS - SdT$$

Since by the Thermodynamic identity, $dU - TdS = - PdV$:

$$\implies dF = -PdV - SdT$$

My textbook then notes that given the form of the above equation, $F$'s natural variables are $V$ and $T$. However, my last equation was just a rewriting of $dF = dU - TdS - SdT$ and is an equally valid equation for $F$.

However, given the form of my most recent equation I'd say $F$'s natural variables are $U$, $S$ and $T$. Why is this wrong?

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Playing with differentials in the way you report, can be done but with some care about the mathematical meaning of the manipulation.

When we write a differential, we should know in advance which are the independent variables the function we differentiate depends on and there is no substitute for it coming from formal manipulations.

Actually, the issue of the independent variables $F$ depends on, has to be solved much before writing differentials: it is the definition $F=U-TS$ which is there for the purpose of encoding the same information contained in $U(S,V)$ into a function of $T=\frac{\partial{U}}{\partial{S}},V$. This is a problem which is solved through a Legendre transform$^*$: $$ F(T,V) = U(S(T,V),V)-TS(T,V) $$ where, I have put explicitly the way each quantity has to be interpreted from the mathematical point of view instead as using the sloppy notation $F=U-TS$ often used in the practice. The function $S(T,V)$ is obtained by using the inverse function theorem in connection with the definition $T=\frac{\partial{U}}{\partial{S}}(S,V)$.

Once all this is clear, one can even use the formal manipulation reported in the textbook, but without possibility of confusion about how many and which variables $F$ depends on.

A different way to see why formal manipulations alone are not enough could be the following. Let's assume that we would define $$ F(S,V) = U(S,V) -ST(S,V) $$ where $T=\frac{\partial{U}}{\partial{S}}(S,V)$. With the same formal manipulation one could "prove" that $F$ is again a function of $T$ and $V$, although we have clarly introduced a function of $S$ and $V$. And only on the basis of such previous knowledge, we can correctly interpret in this case $dT$ as the differential of a function of $S$ and $V$ instead as an independent variable.

($^*$) I am referring to Legendre transforms for sake of simplicity, however, the right tool in thermodynamics is the Legendre-Fenchel transform. The conclusion of the above answer won't be changed by such generalization.

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  • $\begingroup$ What explicitly is $S(T,V)$? That is, what function relates $S$ to $T,V$? $\endgroup$ – sangstar May 2 at 9:29
  • $\begingroup$ @sangstar The explicit form depends on the form of the function $S(U,V)$, i.e. depends on the material. $\endgroup$ – GiorgioP May 2 at 10:25
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If some function $f$ is a function of three variables $x,y,z$, then $x,y,$ and $z$ need to be independent of one another in the sense that changing one while holding the others constant needs to make sense.

This is not the case here. $U$ is not independent of $S$ and $T$ (in fact, it is just proportional to $T$ for an ideal gas), so quantities like $$\left(\frac{\partial F}{\partial T}\right)_{U,S}$$ (the partial derivative of $F$ with respect to $T$ while holding $U$ and $S$ constant) don't make any sense. How can I change $T$ without changing $U$?


A different way to put it is that $F$ and $U$ are not thermodynamic variables but rather thermodynamic potentials. Each potential is a function of some set of thermodynamic variables, chosen from sets of conjugate pairs ($p$ and $V$, $S$ and $T$, $\mu$ and $N$, etc).

Note that $U=U(S,V)$ makes sense because entropy can be changed independently of volume and vice-versa. The same holds for the Helmholtz potential $F=F(T,V)$, the enthalpy $H=H(S,p)$, and the Gibbs potential $G=G(T,p)$.

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  • $\begingroup$ $U$ is just proportional to $T$ only in the case of perfect gases. However, the fomula defining the Helmholtz free energy has much broader validity than for perfect gases. Also making differences between thermodynamic variables and potentials is not clarifying. Think for example $pV$: as a function of $(V,T,\mu)$ it is a thermodynamic potential. $\endgroup$ – GiorgioP May 1 at 21:31
  • $\begingroup$ @GiorgioP Thanks, I fixed the first part. I’m not sure I understand your second point, though. I agree that pV is a valid thermodynamic potential, but my point was that thermodynamic potentials are not functions of one another, but rather are obtained from one another by Legendre transform. $\endgroup$ – J. Murray May 1 at 21:43
  • $\begingroup$ Check out the Born Square - or better Google for the Enhanced Born Square. $\endgroup$ – Cinaed Simson May 1 at 21:45

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